Question: $h(x)=\begin{cases} \dfrac{1}{2x}&\text{for }x\leq-2 \\\\ 2^x&\text{for }-2<x\leq0 \end{cases}$ Find $\lim_{x\to -2}h(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-2$ (Choice B) B $-\dfrac14$ (Choice C) C $\dfrac14$ (Choice D) D The limit doesn't exist.
Solution: $x=-2$ is on the boundary between the pieces of our piecewise function. In order to find $\lim_{x\to -2}h(x)$, we need to find the one-sided limits. Let's find the limit as $x$ approaches $-2$ from the left. We will use the fact that $h(x)=\dfrac{1}{2x}$ for $x$ -values smaller than $-2$. $\begin{aligned} &\phantom{=}\lim_{x\to -2^-}h(x) \\\\ &=\lim_{x\to -2^-}\dfrac{1}{2x} \\\\ &=\dfrac{1}{2(-2)}&\gray{\text{Direct substitution}} \\\\ &=-\dfrac14 \end{aligned}$ Let's find the limit as $x$ approaches $-2$ from the right. We will use the fact that $h(x)=2^x$ for $x$ -values greater than $-2$. $\begin{aligned} &\phantom{=}\lim_{x\to -2^+}h(x) \\\\ &=\lim_{x\to -2^+}2^x \\\\ &=2^{-2}&\gray{\text{Direct substitution}} \\\\ &=\dfrac14 \end{aligned}$ $-\dfrac14\neq \dfrac14$ so the one-sided limits aren't equal. This means that the two-sided limit $\lim_{x\to -2}h(x)$ doesn't exist.